Reverse decimal digits in javascript

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How do I reverse the digits of a number using bitwise?

input:

x = 123;

output:

x = 321;

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14条回答

闹比i

2021-02-05 17:09
That's not inverting bits; that's reversing the order of decimal digits, which is completely different. Here's one way:
```
var x = 123;
var y = 0;
for(; x; x = Math.floor(x / 10)) {
    y *= 10;
    y += x % 10;
}
x = y;
```
If you actually want to invert bits, it's:
```
x = ~x;
```
As a function:
```
function reverse(n) {
    for(var r = 0; n; n = Math.floor(n / 10)) {
        r *= 10;
        r += n % 10;
    }
    return r;
}
```
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时光取名叫无心

2021-02-05 17:11
If you wanted to make a simple reversal:
```
var x = 123;
var y = x.toString();
var z = y.split("").reverse().join("");
var aa = Number(z);
document.write(aa);
```
http://jsfiddle.net/jasongennaro/gV39e/
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情书的邮戳

2021-02-05 17:11

<html>
<script>

function reverseInt(n){

        var r=0;
    while(n!=0){
        r*=10;
        r+=n%10;
        n=Math.floor(n/10);
}
return r;
}


</script>
</html>

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有刺的猬

2021-02-05 17:13

Memory Usage: 35.3 MB, less than 100.00% of JavaScript online submissions for Reverse Integer on leetcode.com. Runtime: 80 ms, faster than 61.48% of JavaScript online submissions for Reverse Integer.

Time complexity is O(log10(n)).

function reverse(x) {
  let rev = 0;
  const isNegative = Math.sign(x) === -1;
  const isOverflow = n => n > 2**31;
  x = Math.abs(x);

  while (x) {
    let pop = x % 10;
    x = Math.floor(x / 10);
    rev = rev * 10 + pop;

    if (isOverflow(rev)) {
      return 0;
    } 
  }

  return isNegative ? rev * -1 : rev;
}

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粉色の甜心

2021-02-05 17:15

//reverse integer
const revInt = (num)=>{
  //turn into string
  if(Math.sign(num)===1)
    return parseInt(num.toString().split('').reverse().join(''));
  else return -1*parseInt(num.toString().split('').reverse().join(''));
}

console.log(revInt(-501));

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不要未来只要你来

2021-02-05 17:16
Reversing The Positive/ Negative Integer Number
```
function reverseInt(n) {
 return parseInt(n.toString().split('').reverse().join()) * Math.sign(n)
}
```
If n is -5, then Math.sign(n)==> will return -1

If n is 5, then Math.sign(n)==> will return 1
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